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Questions like these are best approached with a mix of trial and error and logic.This combines Alternative and Logical approaches.

First, the ones digit of AAB*B is B. This means that B*B is B (because the ones digit of the product is determined only by the ones digits of the factors!).Only three numbers fit the pattern - 0,1, 5 and 6. 0 and 1 are clearly irrelevant so B must be 5 or 6.Next, the tens digits AAB*B is 5 meaning that AB*B has a tens digit of 5.If B=5 then A5*5 has a tens digit of 5 --> 50*A + 25 has a tens of 5 --> 5*A +2 has a ones digit of 5 --> 5*A has a ones digit of 3.This is impossible so B must equal 6.Let's verify:If B=6 then A6*6 has a tens digit of 5 --> 60*A + 36 has a tens of 5 --> 6*A + 3 has a ones digit of 5 --> 6*A has a ones digit of 2 --> A equals 2 or 7.

So, our problem is now 226*6 = C656 or 776*6 = C656Let's solve the first problem as it looks easier. 226*6 = 1200+120+36= 1356. But then B is both 3 and 6. Impossible! Then A must equal 7. 776*6 = 4200 + 420 + 36 = 4656 and C = 4. In this case A+B+C = 7+6+4=17 and (E) is our answer.
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STEP by step approach will help you get answer and that too within decent time..

STEP I - UNIT's digit - Bwhen can be B*B lead to B as units digit..a) \(0*0 = 0\)... But entire answer will become 0b) \(1*1 = 1\)... BUT the product will remain AABc) \(5*5 = 25\)..... But \(5*A+2 = x5\).... Not possible.. the digit cannot be 5, it will be "5*even+2=x2" OR "5*odd+2 = 7"d) \(6*6=36\) ..... \(6*A+3=x5\)... Possible when A is 2 or A is 7no other possibility, so B is 6 and A is 2 or 7

STEP II - A?our product is \(AA6*6=C656\)...if A is 2, it becomes \(226*6 = 1356\) Not same as C656